Question

1. Change the IP multicast address 233.50.18.2 to Ethernet multicast addresses.

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2. Answer the following question about client-server programs:

point-to-point N1 R1 R2

a. Which program is running on the local machine

b. Which program is running on the remote machine (client or server)?

c. Which program is an infinitive program (client or server)?

d. Which program is a finite program (client or server)?

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3. The following is a dump of a UDP header in hexadecimal format: CB820045007C12B2

a. What is the source port number (in decimal)?

b. What is the destination port number (in decimal)?

c. What is the client that uses the services of UDP in this case?

Answer 1

Hi,

3. Given UDP header dump is CB820045007C12B2

a. Source port is given by First four hexadecimal digits which are CB82 which is 52098 in decimal

b.Destination port is given by the next 4 hexadecimal digits which are 0045 which is 69 in decimal

c. Since destination port is 69, Port 69 is the Trivial File Transfer Protocol (TFTP) port. By convention, all machines use these port numbers for these services.

2. Client server programs is a paradigm where client sends requests to server and server is reponsible for processing them.so by definition this paradigm says that client is a finite program which runs in local machine and sends requests to server which is an infinite program running on the remote machine.

a. Client as it is the request sender

b. Server, because it should be up to take requests from clients

c. Server is an infinite program typically which keeps listening to client requests

d.client is a finite program which only sends requests to server.

1. Ethernet addresses have 48 bit address fields for both source and destination and are expressed in hexadecimal in general. and for multicast the first 24 bits are reserved as 0100 . 5e, followed by 0, and the remaning 23(48-24-1) give the multicast address as below

Given IP multicast is 230 . 50 .18. 2
writing it in decimal

11101001 . 00110010 . 00010010 . 00000010

we have to map this 28 bits onto the 23 bits available which is done by choosing the lower order bits i.e

so the next 23bits are,   0110010 . 00010010 . 00000010 (23 bits from right)

we know the 1st 25 bits are 0100 . 5e and 0 i.e 0001 . 0000 . 0101 . 1110 . 0

now combining both we get

00010000 . 0101 1110 . 00110010 .00010010 . 00000010
which is 0100:5E32:1102

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